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# 数学代写|线性代数代写Linear algebra代考|MA2210 Orthogonal projections and orthogonal decompositions

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## 数学代写|线性代数代写Linear algebra代考|Orthogonal projections and orthogonal decompositions

We say that two vectors $v, w$ in an inner product space are orthogonal if $v \cdot w=0$.
Definition 7.1 Let $V$ be a real inner product space, and $U$ a subspace of $V$. The orthogonal complement of $U$ is the set of all vectors which are orthogonal to everything in $U$ :
$$U^{\perp}={w \in V: w \cdot u=0 \text { for all } u \in U}$$
Proposition 7.1 If $V$ is an inner product space and $U$ a subspace of $V$, with $\operatorname{dim}(V)=n$ and $\operatorname{dim}(U)=r$, then $U^{\perp}$ is a subspace of $V$, and $\operatorname{dim}\left(U^{\perp}\right)=n-r$. Moreover, $V=U \oplus U^{\perp}$.

Proof Proving that $U^{\perp}$ is a subspace is straightforward from the properties of the inner product. If $w_1, w_2 \in U^{\perp}$, then $w_1 \cdot u=w_2 \cdot u=0$ for all $u \in U$, so $\left(w_1+w_2\right) \cdot u=0$ for all $u \in U$, whence $w_1+w_2 \in U^{\perp}$. The argument for scalar multiples is similar.

## 数学代写|线性代数代写Linear algebra代考|The Spectral Theorem

The main theorem can be stated in two different ways. I emphasise that these two theorems are the same! Either of them can be referred to as the Spectral Theorem.
Theorem 7.4 If $\alpha$ is a self-adjoint linear map on a real inner product space $V$, then the eigenspaces of $\alpha$ form an orthogonal decomposition of $V$. Hence there is an orthonormal basis of $V$ consisting of eigenvectors of $\alpha$. Moreover, there exist orthogonal projections $\pi_1, \ldots, \pi_r$ satisfying $\pi_1+\cdots+\pi_r=I$ and $\pi_i \pi_j=O$ for $i \neq j$, such that
$$\alpha=\lambda_1 \pi_1+\cdots+\lambda_r \pi_r$$
where $\lambda_1, \ldots, \lambda_r$ are the distinct eigenvalues of $\alpha$.
Theorem 7.5 Let A be a real symmetric matrix. Then there exists an orthogonal matrix $P$ such that $P^{-1} A P$ is diagonal. In other words, any real symmetric matrix is orthogonally similar to a diagonal matrix.

Proof The second theorem follows from the first, since the transition matrix from one orthonormal basis to another is an orthogonal matrix. So we concentrate on the first theorem. It suffices to find an orthonormal basis of eigenvectors, since all the rest follows from our remarks about projections, together with what we already know about diagonalisable maps.

The proof will be by induction on $n=\operatorname{dim}(V)$. There is nothing to do if $n=1$. So we assume that the theorem holds for $(n-1)$-dimensional spaces.
The first job is to show that $\alpha$ has an eigenvector.
Choose an orthonormal basis; then $\alpha$ is represented by a real symmetric ma$\operatorname{trix} A$. Its characteristic polynomial has a root $\lambda$ over the complex numbers. (The so-called “Fundamental Theorem of Algebra” asserts that any polynomial over $\mathbb{C}$ has a root.) We temporarily enlarge the field from $\mathbb{R}$ to $\mathbb{C}$. Now we can find a column vector $v \in \mathbb{C}^n$ such that $A v=\lambda v$. Taking the complex conjugate, remembering that $A$ is real, we have $A \bar{v}=\bar{\lambda} \bar{v}$.
If $v=\left[\begin{array}{llll}z_1 & z_2 & \cdots & z_n\end{array}\right]^{\top}$, then we have
\begin{aligned} \bar{\lambda}\left(\left|z_1\right|^2+\left|z_2\right|^2+\cdots+\left|z_n\right|^2\right) & =\bar{\lambda} \bar{v}^{\top} v \ & =(A \bar{v})^{\top} v \ & =\bar{v}^{\top} A v \ & =\bar{v}^{\top}(\lambda v) \ & =\lambda\left(\left|z_1\right|^2+\left|z_2\right|^2+\cdots+\left|z_n\right|^2\right) \end{aligned}
so $(\bar{\lambda}-\lambda)\left(\left|z_1\right|^2+\left|z_2\right|^2+\cdots+\left|z_n\right|^2\right)=0$. Since $v$ is not the zero vector, the second factor is positive, so we must have $\bar{\lambda}=\lambda$, that is, $\lambda$ is real.

## 数学代写线性代数代写Linear algebra代考|Orthogonal projections and orthogonal decompositions

$$U^{\perp}=w \in V: w \cdot u=0 \text { for all } u \in U$$

## 数学代写|线性代数代写Linear algebra代考|The Spectral Theorem

$$\alpha=\lambda_1 \pi_1+\cdots+\lambda_r \pi_r$$

$$\bar{\lambda}\left(\left|z_1\right|^2+\left|z_2\right|^2+\cdots+\left|z_n\right|^2\right)=\bar{\lambda} \bar{v}^{\top} v \quad=(A \bar{v})^{\top} v=\bar{v}^{\top} A v \quad=\bar{v}^{\top}(\lambda v)=\lambda\left(\left|z_1\right|^2+\left|z_2\right|^2+\cdots+\left|z_n\right|^2\right)$$

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